Probability Ques 76
For a student to qualify, he must pass atleast two out of three exams. The probability that he will pass the 1st exam is $p$. If he fails in one of the exams, then the probability of his passing in the next exam, is $\frac{p}{2}$ otherwise it remains the same. Find the probability that he will qualify.
(2003, 2M)
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Answer:
Correct Answer: 76.$(2 p^{2}-p^{3})$
Solution:
Formula:
- Let $E _i$ denotes the event that the students will pass the $i$ th exam, where $i=1,2,3$
and $E$ denotes the student will qualify.
$\therefore \quad P(E)= {\left[P\left(E _1\right) \times P\left(E _2 / E _1\right)\right] } $ $ +\left[P\left(E _1\right) \times P\left(E _2^{\prime} / E _1\right) \times P\left(E _3 / E _2^{\prime}\right)\right] $
$ \quad+\left[P\left(E _1^{\prime}\right) \times P\left(E _2 / E^{\prime}{ } _1\right) \times P\left(E _3 / E _2\right)\right] $
$ =p^{2}+p(1-p) \cdot \frac{p}{2}+(1-p) \cdot \frac{p}{2} \cdot p $
$\Rightarrow \quad P(E)=\frac{2 p^{2}+p^{2}-p^{3}+p^{2}-p^{3}}{2}=2 p^{2}-p^{3}$
BETA
